Question

In a meeting of mimes, mime 1 goes through a displacement${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{j}}$and mime 2 goes through a displacement${\overrightarrow{\mathbf{d}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\stackrel{\mathit{⏜}}{i}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{j}}$. What are (a) ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{\times }{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$, (b) ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{.}{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$, (c) $({\overrightarrow{d}}_1+{\overrightarrow{d}}_2){\overrightarrow{d}}_2$, and (d) the component of${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$along the direction of${\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$? (Hint: For (d), see Eq.3-20and fig3-18.)

Short answer

(a) The cross product ${\overrightarrow{\mathrm{d}}}_1\times {\overrightarrow{\mathrm{d}}}_2$is $31\stackrel{⏜}{k}$.

(b) The dot product ${\overrightarrow{\mathrm{d}}}_1.{\overrightarrow{\mathrm{d}}}_2$is 8.0

(c) The vector operation role="math" localid="1657946010899" $\mathit{(}{\overrightarrow{\mathit{d}}}_{\mathit{1}}\mathit{+}{\overrightarrow{\mathit{d}}}_{\mathit{2}}\mathit{)}{\overrightarrow{d}}_{\mathit{2}}$is equal to 33

(d) The component of ${\overrightarrow{\mathit{d}}}_{\mathit{1}}$along the direction of ${\overrightarrow{d}}_2$is 1.6

Step-by-step solution

Given data

The vectors are given below:

$$\begin{gathered} {\overrightarrow{d}}_1=4.0\stackrel{⏜}{i}+5.01\stackrel{⏜}{j} \\ {\overrightarrow{d}}_2=-3.0\stackrel{⏜}{i}+4.0\stackrel{⏜}{j} \end{gathered}$$

Understanding the concept

Usethe rules of vector product, dot product, and cross product. The dot product of two vectors is a scalar quantity and the cross product of two vectors is a vector quantity.

The angle between the vectors can be calculated as,

$\cos \theta =\frac{\left(\overrightarrow{d_1}.{\overrightarrow{d}}_2\right)}{\left(d_1.d_2\right)}$ (i)

The cross product is calculated as,

$\overrightarrow{A}\times \overrightarrow{B}=\stackrel{⏜}{i}\left(A_y{B}_z-A_z{B}_y\right)-\stackrel{⏜}{j}\left(A_x{B}_z-A_z{B}_x\right)+\stackrel{⏜}{k}\left(A_x{B}_y-A_y{B}_x\right)$ (ii)

Calculate d→1×d→2

Given vectors do not have $\stackrel{⏜}{k}$components, so $\stackrel{⏜}{i}$and $\stackrel{⏜}{j}$components of the cross product would be zero. Hence, we have only $\stackrel{⏜}{k}$component in the cross product. The cross product can be found using equation (ii) as follows:

$$\begin{gathered} {\overrightarrow{d}}_1\times {\overrightarrow{d}}_2=\left(4\stackrel{⏜}{i}+5.01\stackrel{⏜}{j}\right)\times \left(-3\stackrel{⏜}{i}+04\stackrel{⏜}{j}\right) \\ =\stackrel{⏜}{i}\left[\left(5\times 0\right)-\left(0\times 4\right)\right]-\stackrel{⏜}{j}\left[\left(4\times 0\right)-\left(0\times -3\right)\right]+\stackrel{⏜}{k}\left[\left(4\times 4\right)-\left(-3\times 5\right)\right] \\ =\stackrel{⏜}{i}\left[0\right]-\stackrel{⏜}{j}\left[0\right]+\stackrel{⏜}{k}\left[\left(16\right)-\left(-15\right)\right] \\ =31\stackrel{⏜}{k} \end{gathered}$$

Therefore, role="math" localid="1657946771878" ${\overrightarrow{d}}_1\times {\overrightarrow{d}}_2$is $31\stackrel{⏜}{k}$.

(b) Calculate d→1.d→2

Use the dot product formula to calculate the dot product${\overrightarrow{d}}_1.{\overrightarrow{d}}_2$.

role="math" localid="1657946958877" $$\begin{gathered} {\overrightarrow{d}}_1.{\overrightarrow{d}}_2=\left(4\stackrel{⏜}{i}+5.01\stackrel{⏜}{j}\right).\left(-3\stackrel{⏜}{i}+04\stackrel{⏜}{j}\right) \\ =8.0 \end{gathered}$$)

Therefore, dot product role="math" localid="1657946883238" ${\overrightarrow{d}}_1.{\overrightarrow{d}}_2$is 8.0 .

(c) Calculate (d1+d2).d2

Now, to calculate the $\left({\overrightarrow{d}}_1.{\overrightarrow{d}}_2\right).{\overrightarrow{d}}_2$, first simply the equation as follows.

$\left({\overrightarrow{d}}_1.{\overrightarrow{d}}_2\right).{\overrightarrow{d}}_2={\overrightarrow{d}}_1.{\overrightarrow{d}}_2+d_2^2$

Now, calculate$d_2^2$ .

$$\begin{gathered} d_2^2={\left(-3\right)}^2+{\left(4\right)}^2 \\ =25 \end{gathered}$$

Use the value of ${\overrightarrow{d}}_1.{\overrightarrow{d}}_2$from step (4) to calculate $\left({\overrightarrow{d}}_1.{\overrightarrow{d}}_2\right){\overrightarrow{.d}}_2$

$$\begin{gathered} \left({\overrightarrow{d}}_1.{\overrightarrow{d}}_2\right){\overrightarrow{.d}}_2=8+25 \\ =33 \end{gathered}$$

(d) Calculate the component of d1 along the direction of d2  

The magnitude of ${\overrightarrow{d}}_1$is as follows:

$$\begin{gathered} \left|d_1\right|=\sqrt{4^2+5^2} \\ =6.4\mathrm{m} \end{gathered}$$

Use equation (i) to find the direction.

$$\begin{gathered} \cos \theta =\frac{{\overrightarrow{d}}_1.{\overrightarrow{d}}_1}{d_1{d}_2} \\ \theta ={\cos }^{-1}\left(\frac{\left(6.4\right).\left(5\right)}{8}\right) \\ =75.5° \end{gathered}$$

So horizontal component of${\overrightarrow{d}}_1$ is as follows:

$$\begin{gathered} {\overrightarrow{d}}_{1horizontal}=6.4.\cos \left(75.5\right) \\ =1.6 \end{gathered}$$

Therefore, the horizontal component of ${\overrightarrow{d}}_1$is equal to 1.6 .